## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 10 - Cumulative Review Exercises - Page 1127: 19

#### Answer

The value of the expression $2{{x}^{3}}+3{{x}^{2}}-8x+3=0$ is $\left\{ -3,\frac{1}{2},1 \right\}$

#### Work Step by Step

We have to find the value of x from $2{{x}^{3}}+3{{x}^{2}}-8x+3=0$ By the trial method, $x=1$ is a solution of the equation. So, $x-1$ is a factor of the given equation. Then, divide $2{{x}^{3}}+3{{x}^{2}}-8x+3=0$ by $x-1$: $\frac{\left( 2{{x}^{3}}+3{{x}^{2}}-8x+3 \right)}{x-1}=2{{x}^{2}}+5x-3$ We have obtained a quadratic equation. Now, obtain the factors of $2{{x}^{2}}+5x-3$ as given below. \begin{align} & 2{{x}^{2}}+5x-3=0 \\ & 2{{x}^{2}}+6x-x-3=0 \\ & 2x\left( x+3 \right)-1\left( x+3 \right)=0 \\ & \left( 2x-1 \right)\left( x+3 \right)=0 \end{align} So, $x=\frac{1}{2}$ and $x=-3$. Hence, the values of x are $\left\{ -3,\frac{1}{2},1 \right\}$.

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